package leetcode;

import java.util.PriorityQueue;

/**
 * 	Given two arrays of length m and n with digits 0-9 representing 
 * 	two numbers. Create the maximum number of length k <= m + n from digits of the two. 
 * 	The relative order of the digits from the same array must be preserved. 
 * 	Return an array of the k digits. You should try to optimize your time and space complexity.
 *	
 *	Example 1:
 *	nums1 = [3, 4, 6, 5]
 *	nums2 = [9, 1, 2, 5, 8, 3]
 *	k = 5
 *	return [9, 8, 6, 5, 3]
 *	
 *	Example 2:
 *	nums1 = [6, 7]
 *	nums2 = [6, 0, 4]
 *	k = 5
 *	return [6, 7, 6, 0, 4]
 *	
 *	Example 3:
 *	nums1 = [3, 9]
 *	nums2 = [8, 9]
 *	k = 3
 *	return [9, 8, 9]
 */

/**
 * @author      zxwtry
 * @email       zxwtry@qq.com
 * @project     OJ
 * @package     leetcode
 * @file        P321_CreateMaximumNumber.java
 * @type        P321_CreateMaximumNumber
 * @date        2017年1月9日 下午9:07:53
 * @details     Solution2: AC 25 46.78%
 */
public class P321_CreateMaximumNumber {
	static class Solution1 {
	    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
	        if (nums1 == null || nums2 == null || k < 0 || k > nums1.length + nums2.length)
	        	return new int[0];
	        PriorityQueue<T> priorityQueue = new PriorityQueue<T>();
	        for (int num : nums1) priorityQueue.add(new T(num));
	        for (int num : nums2) priorityQueue.add(new T(num));
	        int[] maxNumber = new int[k];
	        for (int maxNumberIndex = 0; maxNumberIndex < maxNumber.length; maxNumberIndex ++)
	            maxNumber[maxNumberIndex] = priorityQueue.poll().val;
	        return maxNumber;
	    }
	    private static class T implements Comparable<T> {
	        int val;
	        T(int val) {
	            this.val = val;
	        }
            @Override
            public int compareTo(T o) {
                return Integer.compare(o.val, this.val);
            }
	    }
	}
	static class Solution2 {
	    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
	        int n = nums1.length;
	        int m = nums2.length;
	        int[] ans = new int[k];
	        for (int i = Math.max(0, k - m); i <= k && i <= n; ++i) {
	            int[] candidate = merge(maxArray(nums1, i), maxArray(nums2, k - i), k);
	            if (greater(candidate, 0, ans, 0)) ans = candidate;
	        }
	        return ans;
	    }
	    private int[] merge(int[] nums1, int[] nums2, int k) {
	        int[] ans = new int[k];
	        for (int i = 0, j = 0, r = 0; r < k; ++r)
	            ans[r] = greater(nums1, i, nums2, j) ? nums1[i++] : nums2[j++];
	        return ans;
	    }
	    public boolean greater(int[] nums1, int i, int[] nums2, int j) {
	        while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) {
	            i++;
	            j++;
	        }
	        return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);
	    }
	    public int[] maxArray(int[] nums, int k) {
	        int n = nums.length;
	        int[] ans = new int[k];
	        for (int i = 0, j = 0; i < n; ++i) {
	            while (n - i + j > k && j > 0 && ans[j - 1] < nums[i]) j--;
	            if (j < k) ans[j++] = nums[i];
	        }
	        return ans;
	    }
	}
}
